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Question

In ∆ABC, ∠B = 35°, ∠C = 65° and the bisector AD of ∠BAC meets BC at D. Then, which of the following is true?
(a) AD > BD > CD
(b) BD > AD > CD
(c) AD > CD > BD
(d) None of these

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Solution

(b)BD>AD>CD

In ABC, we have:
B=35°,C=65° and the bisector AD of BAC meets BC at D.
Then we have:
A+B+C=180°A+35°+65°=180°A=80°

AD is the angle bisector of BAC
BAD=CAD=40°

Now, in triangle ABD, we have:
BAD>ABD
BD>AD
Also, in triangle ACD, we have:
ACD>CAD
AD>CD
BD>AD>CD

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